Evaluate $\int \left( \frac { x } { 9 } - 9 \right) ^ { 2 / 9 } d x$ .
A) $\left( \frac { x } { 9 } - 9 \right) ^ { 11 / 9 } + C$
B) $\frac { 9 } { 11 } \left( \frac { x } { 9 } - 9 \right) ^ { 11 / } + C$
C) $\frac { 81 } { 11 } \left( \frac { x } { 9 } - 9 \right) ^ { 11 / 9 } + C$
D) $\frac { x ^ { 2 } } { 18 } - 9 x + C$

Question

Quizplus · Accepted Answer

To solve the integral $\int \left( \frac { x } { 9 } - 9 \right) ^ { 2 / 9 } d x$, we use the substitution method. Let $u = \frac{x}{9} - 9$, then $du = \frac{1}{9}dx$, or $dx = 9du$. Substituting these into the integral gives us $\int u^{2/9} \cdot 9 du$, which simplifies to $9 \int u^{2/9} du$. Integrating $u^{2/9}$ with respect to $u$ gives $\frac{9}{11}u^{11/9} + C$, substituting back for $u$ gives $\frac{81}{11}\left(\frac{x}{9} - 9\right)^{11/9} + C$.

Evaluate ∫(x9−9)2/9dx\int \left( \frac { x } { 9 } - 9 \right) ^ { 2 / 9 } d x∫(9x​−9)2/9dx

Evaluate $\int \left( \frac { x } { 9 } - 9 \right) ^ { 2 / 9 } d x$